作者:遗忘的睡骨 | 来源:互联网 | 2023-01-29 13:45
我试图弄清楚如何为此构建工作流sklearn.neighbors.KNeighborsRegressor
,包括:
归一化特征
特征选择(20个数字特征的最佳子集,无特定总数)
在1到20的范围内交叉验证超参数K
交叉验证模型
使用RMSE作为误差指标
scikit-learn中有很多不同的选项,我在决定我需要的类时有点不知所措。
此外sklearn.neighbors.KNeighborsRegressor
,我认为我需要:
sklearn.pipeline.Pipeline
sklearn.preprocessing.Normalizer
sklearn.model_selection.GridSearchCV
sklearn.model_selection.cross_val_score
sklearn.feature_selection.selectKBest
OR
sklearn.feature_selection.SelectFromModel
有人可以告诉我定义此管道/工作流程的样子吗?我认为应该是这样的:
import numpy as np
from sklearn.pipeline import Pipeline
from sklearn.preprocessing import Normalizer
from sklearn.feature_selection import SelectKBest, f_classif
from sklearn.neighbors import KNeighborsRegressor
from sklearn.model_selection import cross_val_score, GridSearchCV
# build regression pipeline
pipeline = Pipeline([('normalize', Normalizer()),
('kbest', SelectKBest(f_classif)),
('regressor', KNeighborsRegressor())])
# try knn__n_neighbors from 1 to 20, and feature count from 1 to len(features)
parameters = {'kbest__k': list(range(1, X.shape[1]+1)),
'regressor__n_neighbors': list(range(1,21))}
# outer cross-validation on model, inner cross-validation on hyperparameters
scores = cross_val_score(GridSearchCV(pipeline, parameters, scoring="neg_mean_squared_error", cv=10),
X, y, cv=10, scoring="neg_mean_squared_error", verbose=2)
rmses = np.abs(scores)**(1/2)
avg_rmse = np.mean(rmses)
print(avg_rmse)
它似乎没有出错,但是我的一些担忧是:
我是否正确执行了嵌套的交叉验证,以使我的RMSE不受偏见?
如果我想最终的模型根据最佳RMSE进行选择,我应该用scoring="neg_mean_squared_error"
两个cross_val_score
和GridSearchCV
?
是SelectKBest, f_classif
用于选择KNeighborsRegressor
模型特征的最佳选择吗?
我怎么看:
哪个功能子集被选为最佳
哪个K被选为最佳
任何帮助是极大的赞赏!
1> makis..:
您的代码似乎还可以。
对于scoring="neg_mean_squared_error"
两个cross_val_score
和GridSearchCV
,我会做同样的,以确保一切正常,可测试这一点的唯一方法是删除这两个中的一个,看看结果的变化。
SelectKBest
是一个很好的方法,但您也可以使用此处SelectFromModel
找到的其他方法
最后,为了获得最佳的参数和功能得分,我对您的代码进行了一些修改,如下所示:
import ...
pipeline = Pipeline([('normalize', Normalizer()),
('kbest', SelectKBest(f_classif)),
('regressor', KNeighborsRegressor())])
# try knn__n_neighbors from 1 to 20, and feature count from 1 to len(features)
parameters = {'kbest__k': list(range(1, X.shape[1]+1)),
'regressor__n_neighbors': list(range(1,21))}
# changes here
grid = GridSearchCV(pipeline, parameters, cv=10, scoring="neg_mean_squared_error")
grid.fit(X, y)
# get the best parameters and the best estimator
print("the best estimator is \n {} ".format(grid.best_estimator_))
print("the best parameters are \n {}".format(grid.best_params_))
# get the features scores rounded in 2 decimals
pip_steps = grid.best_estimator_.named_steps['kbest']
features_scores = ['%.2f' % elem for elem in pip_steps.scores_ ]
print("the features scores are \n {}".format(features_scores))
feature_scores_pvalues = ['%.3f' % elem for elem in pip_steps.pvalues_]
print("the feature_pvalues is \n {} ".format(feature_scores_pvalues))
# create a tuple of feature names, scores and pvalues, name it "features_selected_tuple"
featurelist = ['age', 'weight']
features_selected_tuple=[(featurelist[i], features_scores[i],
feature_scores_pvalues[i]) for i in pip_steps.get_support(indices=True)]
# Sort the tuple by score, in reverse order
features_selected_tuple = sorted(features_selected_tuple, key=lambda
feature: float(feature[1]) , reverse=True)
# Print
print 'Selected Features, Scores, P-Values'
print features_selected_tuple
使用我的数据的结果:
the best estimator is
Pipeline(steps=[('normalize', Normalizer(copy=True, norm='l2')), ('kbest', SelectKBest(k=2, score_func=)), ('regressor', KNeighborsRegressor(algorithm='auto', leaf_size=30, metric='minkowski',
metric_params=None, n_jobs=1, n_neighbors=18, p=2,
weights='uniform'))])
the best parameters are
{'kbest__k': 2, 'regressor__n_neighbors': 18}
the features scores are
['8.98', '8.80']
the feature_pvalues is
['0.000', '0.000']
Selected Features, Scores, P-Values
[('correlation', '8.98', '0.000'), ('gene', '8.80', '0.000')]